Mean squared error of the sample maximum

Question

A machine's output is uniformly distributed between 0 and an unknown upper limit $\theta$ . From $n = 4$ independent readings you estimate $\theta$ by the largest reading, $M = \max(X_1, \dots, X_4)$ . Take $\theta = 10$ .

What is the mean squared error of $M$ ?

Accepted Answer

The idea. The maximum of a sample can never exceed the true ceiling 	heta, so it is always biased low. Its MSE combines that downward bias with its scatter. The two moments of the maximum. For n uniform draws on (0,	heta), \mathbb{E}[M] = \frac{n}{n+1}	heta, \qquad 	ext{Var}(M) = \frac{n}{(n+1)^2(n+2)}	heta^2. The bias is \mathbb{E}[M] - 	heta = -	heta/(n+1): the max sits, on average, one "gap" below the ceiling. Combine into MSE. Adding variance and squared bias, the pieces collapse neatly: 	ext{MSE}(M) = \frac{n}{(n+1)^2(n+2)}	heta^2 + \frac{1}{(n+1)^2}	heta^2 = \frac{2	heta^2}{(…

Mean squared error of the sample maximum

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